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julia-pushkina [17]
3 years ago
10

Please help me Simplify the expression -2x^2(4x − 3)

Mathematics
1 answer:
miv72 [106K]3 years ago
6 0
-2x^2(4x − 3)
=-8x^3 + 6x^2

hope it helps

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den301095 [7]
I think we need to divide some numbers there
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3 years ago
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What are the coordinates of the vertex of the parabola described by the equation below? y=7(x+5)^2-4
Ipatiy [6.2K]

Answer:

y= 7(x+5)^2 -4

The vertex form for a parabola is given by this expression:

y = a(x-h)^2 +k

By direct comparison we see that for this case:

a = 7, h = -5 and k=-4

And we know from the general expression that the vertex is:

V= (h,k)

So then the vertex for this case is:

V= (-5,-4)

Step-by-step explanation:

For this case we have the following function:

y= 7(x+5)^2 -4

And we need to take in count that the vertex form for a parabola is given by this expression:

y = a(x-h)^2 +k

By direct comparison we see that for this case:

a = 7, h = -5 and k=-4

And we know from the general expression that the vertex is:

V= (h,k)

So then the vertex for this case is:

V= (-5,-4)

4 0
2 years ago
A regular octagon has a perimeter of 48 cm. What is the length of each side?
Anarel [89]

Answer:

Each side of the octagon a length of 6 cm

Step-by-step explanation:

Divide the perimeter by the amount of sides the shape has.

48 ÷ 8 = 6

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3 years ago
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Whattttt minion meme
AVprozaik [17]

Answer:

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Step-by-step explanation:

5 0
3 years ago
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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by
a_sh-v [17]

Answer:

Dimension of the box is 16.1\times 7.1\times 2.45

The volume of the box is 280.05 in³.

Step-by-step explanation:          

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the box?

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is L=21-2h

The width of the box is W=12-2h

The volume of the box is V=L\times W\times H

V=(21-2h)\times (12-2h)\times h

V=(21-2h)\times (12h-2h^2)

V=252h-42h^2-24h^2+4h^3

V=4h^3-66h^2+252h

To maximize the volume we find derivative of volume and put it to zero.

V'=12h^2-132h+252

0=12h^2-132h+252

Solving by quadratic formula,

h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}

h=\frac{132\pm72.99}{24}

h=2.45,8.54

Now, substitute the value of h in the volume,

V=4h^3-66h^2+252h

When, h=2.45

V=4(2.45)^3-66(2.45)^2+252(2.45)

V\approx 280.05

When, h=8.54

V=4(8.54)^3-66(8.54)^2+252(8.54)

V\approx -170.06

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is L=21-2(2.45)=16.1

The width of the box is W=12-2(2.45)=7.1

So, Dimension of the box is 16.1\times 7.1\times 2.45

6 0
3 years ago
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